The number of solutions of the equation x + 2 | vedclass

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(A) To find the number of solutions for $x + 2 	an x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$,we rewrite the equation as:$2 	an x = \frac{\pi}{2

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$3$

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(A) To find the number of solutions for $x + 2 	an x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$,we rewrite the equation as:$2 	an x = \frac{\pi}{2} - x$$	an x = -\frac{1}{2}x + \frac{\pi}{4}$We look for the intersection points of the graphs $y = 	an x$ and $y = -\frac{1}{2}x + \frac{\pi}{4}$ within the interval $[0, 2\pi]$.$1$. In the interval $[0, \frac{\pi}{2})$,$	an x$ increases from $0$ to $\infty$,while the line $y = -\frac{1}{2}x + \frac{\pi}{4}$ decreases from $\frac{\pi}{4} \approx 0.785$ to $0$. There is exactly $1$ intersection point.$2$. In the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$,$	an x$ increases from $-\infty$ to $\infty$,while the line decreases from $0$ to $-\frac{\pi}{2} \approx -1.57$. There is exactly $1$ intersection point.$3$. In the interval $(\frac{3\pi}{2}, 2\pi]$,$	an x$ increases from $-\infty$ to $0$,while the line decreases from $-\frac{\pi}{2}$ to $-\frac{3\pi}{4} \approx -2.35$. There is exactly $1$ intersection point.Thus,there are a total of $3$ solutions.

The number of solutions of the equation $x + 2 \tan x = \frac{\pi}{2}$ in the interval $[0, 2\pi]$ is:

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